Ncert Exercises & Solutions

Rate law for the reaction $A+2 B \rightarrow C$ is found to be
Rate = k[A][B]
If the concentration of reactant 'B' is doubled, keeping the concentration of A constant, then the value of the rate of the reaction will be:

1.	The same.	2.	Doubled.
3.	Quadrupled.	4.	Halved.

HINT: Rate of reaction depends on the concentration of reactant.

Explanation:

Rate law can be written as

Rate = k[A][B]

The rate of reaction w.r.t B is of the first order.

$R_{1}$ = k[A][B]

When the concentration of reactant B' is doubled then calculate $R_{2}$ as follows:

$\begin{array}{l} R_{1} = k [A] [B] \\ R_{2} = k [A] [2 B] \\ \frac{R_{1}}{R_{2}} = \frac{1}{2} \\ R_{2} = 2 R_{1} \end{array}$

Therefore, as the concentration of B is doubled keeping the concentration of A constant rate of reaction became doubled.

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